# Can the mean of a frequency distribution be regarded as [closed]

Can the mean of a frequency distribution be regarded as a weighted mean with relative frequencies as weights ?

Surely. Let's imagine $$j$$ distinct values $$x_j$$ with frequency weights $$w_j$$. Then a weighted mean is $$\sum_j w_j x_j / \sum_j w_j$$.
Alternatively we can just write down all the values, say $$n$$ in number, whether distinct or not, so each has frequency $$1$$. The mean is then $$(x_1 + \cdots + x_n)/n$$, which we could redundantly write $$(1\ x_1 + \cdots + 1\ x_n) / n$$. This is exactly equivalent to weights $$w_i$$ all $$1$$, yielding expressions that all mean the same for frequencies of $$1$$: $$\sum_{i=1}^n w_i x_i\,/\,\sum_{i=1}^n w_i\ =\ \sum_{i=1}^n w_i x_i\, /\, n\ =\ (\sum_{i=1}^n 1\ x_i)\, /\, n = \sum_{i=1}^n x_i\, /\, n.$$